Not just trigonometry, its graph theory
Let be a trapezoid with with side lengths shown above. . and intersect at right angles.
Find the area of .
Give your answer to 3 decimal places.
The answer is 46.666.
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1 solution
Let the point of intersection of AC and BD be E .
△ABE and △CDE are similar , so AE = (2/5)CE and BE = (2/5)DE .
△AED is a right triangle , so AD^2 = AE^2 + DE^2 = (4/25)CE^2 + DE^2 .
△BEC is a right triangle , so BC^2 = BE^2 + CE^2 = (4/25)DE^2 + CE^2 .
So AD^2 + BC^2 = (29/25)CE^2 + (29/25)DE^2 .
△CED is a right triangle ,
CE^2 + DE^2 = CD^2 = 100
(29/25)CE^2 + (29/25)DE^2 = (29/25)*100 = 116
So AD^2 + BC^2 = 116 .
We know △AQB and △DQC are similar ,
QD = (5/2)QA , so
AD = QD - QA = (3/2)QA
QC = (5/2)QB , so
BC = QC - QB = (3/2)QB
Therefore
[(3/2)QA]^2 + [(3/2)QB]^2 = 116
(9/4)QA^2 + (9/4)QB^2 = 116
QA^2 + QB^2 = 464/9
Use cosine law for △AQB ,
AB^2 = QA^2 + QB^2 - 2QA QB cos45°
16 = 464/9 - 2QA QB (1/√2)
(√2)QA*QB = 464/9 - 16 = 320/9
QA*QB = (320/9) / √2 = 320/(9√2)
Let the area of △QAB be Area(△QAB) ,
Area(△QAB) = (1/2) QA QB*sin45°
= 160/(9√2)*(1/√2)
= 80/9
Let the area of △QDC be Area(△QDC) ,
Area(△QDC) = [(5/2)^2]Area(△QAB)
= (25/4)*(80/9)
= 500/9
So the area of the trapezoid ABCD is Area(△QDC) - Area(△QAB) = 420/9 = 140/3 .